       Re: Finding only the real roots

• To: mathgroup at smc.vnet.net
• Subject: [mg113635] Re: Finding only the real roots
• From: Alexei Boulbitch <alexei.boulbitch at iee.lu>
• Date: Fri, 5 Nov 2010 05:13:29 -0500 (EST)

```Hi, Peter,

All your questions have simple answers, but first your should write down correct Mathematica expressions.

NSolve, rather than Nsolve, Sqrt, rather than root and [] at Sqrt, rather than () at root.

You will hardly move further without learning few basic things about Mathematica, so do it.

In particular,
NSolve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p]

{{p ->  (0.125 (-1. (-8. - 8. c - 8. g) g -
11.3137 Sqrt[-0.125 g^2 + 1. c g^2 + 0.5 g^3]))/g^2}, {p ->  (
0.125 (-1. (-8. - 8. c - 8. g) g +
11.3137 Sqrt[-0.125 g^2 + 1. c g^2 + 0.5 g^3]))/g^2}}

but I would do

Solve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p]

{{p ->  (2 g + 2 c g + 2 g^2 - Sqrt[-g^2 + 8 c g^2 + 4 g^3])/(
2 g^2)}, {p ->  (
2 g + 2 c g + 2 g^2 + Sqrt[-g^2 + 8 c g^2 + 4 g^3])/(2 g^2)}}

instead. Also Manipulate works. Try this for instance:

Manipulate[
Solve[1/2 + c + g (1 - p) + Sqrt[c + p*g - 1] == 0, p], {c, 0,
1}, {g, 0, 1}]

Have fun, Alexei

Hi all,
I'm trying to numerically evaluate a function using NSolve, for
example
Nsolve[1/2+c+g(1-p)+root(c+p*g-1)==0,p] where my two variables g and c
are both in the interval [0,1]. But, since the root is not always
positive I do not get a solution to NSolve. The idea was therefore to
first define the positive roots and then to do NSolve.

My questions are therefore;
1) Is this the best approach, or are there more direct ways?
2) What should I do to only get the real roots?
3) If I first solve the root, and then use NSolve, will I still be
able to use functions such as Manipulate on the NSolve solution?

Best,
Petter

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Alexei Boulbitch, Dr. habil.
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